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WarriorWithin
GMAT 1: 650 Q49 V30
Posts: 65
11 Dec 2020, 04:07
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
64% (02:23) correct
36%
(02:33)
wrong
based on 306
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A triangle in the xy-coordinate plane has vertices with coordinates (0, 2), (8, 10), and (15, 0). What is the area of this triangle?
A. 53
B. 68
C. 78
D. 83
E. 100
A. 53
B. 68
C. 78
D. 83
E. 100
sudeshpatodiya
Posts: 89
11 Dec 2020, 05:21
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Quote:
Plot the Triangle on a xy plane.
Attachment:
triangle.png [ 6.39 KiB | Viewed 5334 times ]
As per the image, the are of required triangle ABC = Total area of rectangle OBDE - Sum of Areas of Triangles (AOB + BCD + AEC)
= Length * width + Area of all 3 triangles ( \(\frac{1}{2}\) *Base*Height )
= (10*15) - \(\frac{1}{2}\)(2*15 + 7*10 + 8*8)
= 150 - \(\frac{1}{2}\) (30 + 70 + 64)
= 150 - \(\frac{1}{2}\) (164)
= 150 - 82
= 68
(B)
General Discussion
11 Dec 2020, 05:09
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The area of a triangle given the three vertices is found by the solution of the following determinant. We take the absolute value of the determinant.
\(\frac{1}{2}\)\(\begin{vmatrix}\\
x_1 \space x_2 \space x_3 \\ \\
y_1 \space y_2 \space y_3 \\1 \space \space 1 \space \space 1 \\
\end{vmatrix}\)
The solution to this is \(\frac{1}{2}|[x_1(y_2 - y_3) - x_2(y_1 - y_3) + x_3(y_1 - y_2)]\)|
\(\frac{1}{2}\)\(\begin{vmatrix}\\
0 \space 8 \space 15 \\ \\
2 \space 10 \space 0 \\ 1 \space \space 1 \space \space 1 \\
\end{vmatrix}\)
Therefore the area = \(\frac{1}{2}\) |[0(10 - 0) - 8(2 - 0) + 15(2 - 10)]| = \(\frac{1}{2}\) |[0 - 16 - 120]| = \(\frac{1}{2}\) |-136| = 68
Option B
Arun Kumar
\(\frac{1}{2}\)\(\begin{vmatrix}\\
x_1 \space x_2 \space x_3 \\ \\
y_1 \space y_2 \space y_3 \\1 \space \space 1 \space \space 1 \\
\end{vmatrix}\)
The solution to this is \(\frac{1}{2}|[x_1(y_2 - y_3) - x_2(y_1 - y_3) + x_3(y_1 - y_2)]\)|
\(\frac{1}{2}\)\(\begin{vmatrix}\\
0 \space 8 \space 15 \\ \\
2 \space 10 \space 0 \\ 1 \space \space 1 \space \space 1 \\
\end{vmatrix}\)
Therefore the area = \(\frac{1}{2}\) |[0(10 - 0) - 8(2 - 0) + 15(2 - 10)]| = \(\frac{1}{2}\) |[0 - 16 - 120]| = \(\frac{1}{2}\) |-136| = 68
Option B
Arun Kumar
